Chapter 10 Lecture- Gases

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Chapter 10 Lecture- Gases

  1. 1. Chapter 10- Gases Sections 10.1 - 10.4 Homework assignment #1
  2. 2. Characteristics of Gases <ul><li>Unlike liquids and solids, they </li></ul><ul><ul><li>Expand to fill their containers. (volume of gas = volume of container) </li></ul></ul><ul><ul><li>Are highly compressible. (P  V) </li></ul></ul><ul><ul><li>Have extremely low densities. (much empty space) </li></ul></ul><ul><ul><li>Form homogeneous mixtures with each other. (diffusion) </li></ul></ul><ul><li>Gases can have very different chemical properties, but their physical properties are very similar. </li></ul>
  3. 3. Our atmosphere is a mixture of N 2 (78%) O 2 (21%) Ar (0.9%) others
  4. 5. <ul><li>In gases, the distance between molecules is relatively large. </li></ul>
  5. 6. <ul><li>Pressure is the amount of force applied to an area. </li></ul>Pressure <ul><li>Atmospheric pressure is the weight of air per unit of area. </li></ul>P = F A
  6. 7. Units of Pressure <ul><li>Pascals </li></ul><ul><ul><li>1 Pa = 1 N/m 2 </li></ul></ul><ul><li>Bar </li></ul><ul><ul><li>1 bar = 10 5 Pa = 100 kPa </li></ul></ul>
  7. 8. Units of Pressure <ul><li>mm Hg or torr </li></ul><ul><ul><li>These units are literally the difference in the heights measured in mm ( h ) of two connected columns of mercury. </li></ul></ul><ul><li>Atmosphere </li></ul><ul><ul><li>1.00 atm = 760 torr </li></ul></ul>
  8. 10. <ul><li>The height of the column decreases because atmospheric pressure decreases with increasing altitude. </li></ul>
  9. 11. Manometer <ul><li>Used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel. </li></ul>
  10. 12. What is the pressure of the gas in the bulb? <ul><li>P gas = P h </li></ul><ul><li>P gas = P atm </li></ul><ul><li>P gas = P h + P atm </li></ul><ul><li>P gas = P h - P atm </li></ul><ul><li>P gas = P atm - P h </li></ul>
  11. 13. What is the pressure of the gas in the bulb? <ul><li>P gas = P h </li></ul><ul><li>P gas = P atm </li></ul><ul><li>P gas = P h + P atm </li></ul><ul><li>P gas = P h - P atm </li></ul><ul><li>P gas = P atm - P h </li></ul>
  12. 14. Standard Pressure <ul><li>Normal atmospheric pressure at sea level. </li></ul><ul><li>It is equal to </li></ul><ul><ul><li>1.00 atm </li></ul></ul><ul><ul><li>760 torr (760 mm Hg) </li></ul></ul><ul><ul><li>101.325 kPa </li></ul></ul>
  13. 15. Boyle’s Law <ul><li>The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure. </li></ul>
  14. 16. Boyle’s Law
  15. 17. As P and V are inversely proportional <ul><li>A plot of V versus P results in a curve. </li></ul><ul><li>Since </li></ul><ul><ul><li>V = k (1/ P ) </li></ul></ul><ul><ul><li>This means a plot of V versus 1/ P will be a straight line. </li></ul></ul>PV = k
  16. 20. <ul><li>As you double the pressure, the volume decreases to half its original value. </li></ul>
  17. 21. A gas initially at 2.0 atm is in an adjustable volume container of 10. L in volume. If the pressure is decreased to 0.50 atm, what is the new volume? <ul><li>40. L </li></ul><ul><li>20. L </li></ul><ul><li>10. L </li></ul><ul><li>5.0 L </li></ul>
  18. 22. Correct Answer: P V 1 constant   Thus, PV = constant and P 1 V 1 =P 2 V 2 2.00 atm(10. L) = 0.50 atm ( V final ) V final = 2.00 atm(10. L)/0.50 atm = 40. L <ul><li>40. L </li></ul><ul><li>20. L </li></ul><ul><li>10. L </li></ul><ul><li>5.0 L </li></ul>
  19. 23. Charles’s Law <ul><li>The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. </li></ul>A plot of V versus T will be a straight line. <ul><li>i.e., </li></ul>V T = k
  20. 25. <ul><li>No. The volume decreases but it doesn’t decrease to half because the volume is proportional to temperature on the Kelvin scale (not the Celsius scale). </li></ul>
  21. 26. Assuming pressure is held constant, to what volume will a balloon initially at 1.0 L change if its temperature is decreased from 300 K to 75 K? <ul><li>1.0 L </li></ul><ul><li>2.0 L </li></ul><ul><li>0.25 L </li></ul><ul><li>4.0 L </li></ul>
  22. 27. Correct Answer: T V   constant Thus, T/V = constant and T 1 /V 1 =T 2 /V 2 1.0 L/300 K = ( V final )/75 K V final = 75 K/(1.0 L)300 K = 0.25 L <ul><li>1.0 L </li></ul><ul><li>2.0 L </li></ul><ul><li>0.25 L </li></ul><ul><li>4.0 L </li></ul>
  23. 28. Avogadro’s Law <ul><li>The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. </li></ul><ul><li>Mathematically, this means </li></ul>V = kn
  24. 29. Ideal-Gas Equation <ul><ul><ul><ul><li>V  1/ P (Boyle’s law) </li></ul></ul></ul></ul><ul><ul><ul><ul><li>V  T (Charles’s law) </li></ul></ul></ul></ul><ul><ul><ul><ul><li>V  n (Avogadro’s law) </li></ul></ul></ul></ul><ul><li>So far we’ve seen that </li></ul><ul><li>Combining these, we get </li></ul>V  nT P
  25. 30. Ideal-Gas Equation <ul><li>The constant of proportionality is known as R , the gas constant. </li></ul>
  26. 31. Ideal-Gas Equation <ul><li>The relationship </li></ul>then becomes or PV = nRT nT P V  nT P V = R
  27. 32. Ideal-Gas Equation
  28. 33. At standard temperature and pressure, how many moles of gas are present in a box with a volume of 112 L? <ul><li>1.00 moles </li></ul><ul><li>2.00 moles </li></ul><ul><li>5.00 moles </li></ul><ul><li>0.200 moles </li></ul>
  29. 34. Correct Answer:       L 22.41 atm 1.000 K 273.15 K L·atm/mol· 0.08206 mol 1     P nRT V nRT PV Thus, at STP 22.41 L = 112 L 1.00 mol n n = 5.00 moles <ul><li>1.00 moles </li></ul><ul><li>2.00 moles </li></ul><ul><li>5.00 moles </li></ul><ul><li>0.200 moles </li></ul>
  30. 35. If 250 mL of NO is placed in a flask with O 2 , what volume of O 2 is needed for complete reaction? <ul><li>100 mL </li></ul><ul><li>125 mL </li></ul><ul><li>200 mL </li></ul><ul><li>250 mL </li></ul><ul><li>Cannot be determined from the given information. </li></ul>2 NO ( g ) + O 2 ( g ) 2 NO 2 ( g )
  31. 36. If 250 mL of NO is placed in a flask with O 2 , what volume of O 2 is needed for complete reaction? <ul><li>100 mL </li></ul><ul><li>125 mL </li></ul><ul><li>200 mL </li></ul><ul><li>250 mL </li></ul><ul><li>Cannot be determined from the given information. </li></ul>2 NO ( g ) + O 2 ( g ) 2 NO 2 ( g )
  32. 37. Section 10.4 - 10.6 Second homework assignment
  33. 38. Densities of Gases <ul><li>If we divide both sides of the ideal-gas equation by V and by RT , we get </li></ul>n V P RT =
  34. 39. <ul><li>We know that </li></ul><ul><ul><li>moles  molecular mass = mass </li></ul></ul>Densities of Gases n   = m
  35. 40. <ul><li>We know that </li></ul><ul><ul><li>moles  molecular mass = mass </li></ul></ul>Densities of Gases <ul><li>So multiplying both sides by the molecular mass (  ) gives </li></ul>n   = m P  RT m V =
  36. 41. Densities of Gases <ul><li>Mass  volume = density </li></ul><ul><li>So, </li></ul><ul><li>Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas. </li></ul>P  RT m V = d =
  37. 43. <ul><li>Water vapor is less dense than N 2 . </li></ul>
  38. 44. If an equal mass of each gas is put into a separate balloon, which will have the greatest volume? Assume that they are all the same temperature and pressure. <ul><li>He </li></ul><ul><li>H 2 </li></ul><ul><li>N 2 </li></ul><ul><li>Ne </li></ul><ul><li>O 2 </li></ul>
  39. 45. If an equal mass of each gas is put into a separate balloon, which will have the greatest volume? Assume that they are all the same temperature and pressure. <ul><li>He </li></ul><ul><li>H 2 </li></ul><ul><li>N 2 </li></ul><ul><li>Ne </li></ul><ul><li>O 2 </li></ul>
  40. 46. Molecular Mass <ul><li>We can manipulate the density equation to enable us to find the molecular mass of a gas: </li></ul>Becomes P  RT d = dRT P  =
  41. 47. For two gases under identical conditions, moles are proportional to volume. Stoichiometry can be done with volumes of gases under identical conditions. 2H 2 (g) + O 2 (g)  2H 2 O (g) 2L 1L 2L 44.8L 22.4L 44.8L
  42. 49. N 2 ( g ) + 3 H 2 ( g ) 2 NH 3 ( g ) At STP, 16 L of N 2 and 48 L of H 2 are mixed. Assuming all the reactants are consumed, how many L of NH 3 will be produced? <ul><li>8.0 L </li></ul><ul><li>16 L </li></ul><ul><li>24 L </li></ul><ul><li>32 L </li></ul><ul><li>64 L </li></ul>
  43. 50. Correct Answer: According to Avogadro’s law, mole ratios in the chemical equation will be volume ratios under identical conditions. Because the reactants are in a stoichiometric 3:1 volume ratio, the product will have stoichiometric equivalence. Thus, (16 L N 2 )(2 mol NH 3 /1 mol N 2 ) = 32 L NH 3 ) (constant T P, n V  <ul><li>8.0 L </li></ul><ul><li>16 L </li></ul><ul><li>24 L </li></ul><ul><li>32 L </li></ul><ul><li>64 L </li></ul>
  44. 51. Dalton’s Law of Partial Pressures <ul><li>The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. </li></ul><ul><li>In other words, </li></ul><ul><ul><ul><ul><li>P total = P 1 + P 2 + P 3 + … </li></ul></ul></ul></ul>
  45. 53. <ul><li>The pressure exerted by N 2 gas does not change when O 2 is added to the container. </li></ul>
  46. 54. Solution Because each gas behaves independently, we can use the ideal-gas equation to calculate the pressure that each would exert if the other were not present. The total pressure is the sum of these two partial pressures. SAMPLE EXERCISE 10.10 Applying Dalton’s Law of Partial Pressures A gaseous mixture made from 6.00 g O 2 and 9.00 g CH 4 is placed in a 15.0-L vessel at 0°C. What is the partial pressure of each gas, and what is the total pressure in the vessel? Solve:  We must first convert the mass of each gas to moles: We can now use the ideal-gas equation to calculate the partial pressure of each gas: According to Dalton’s law (Equation 10.12), the total pressure in the vessel is the sum of the partial pressures:
  47. 55. Partial Pressures <ul><li>When one collects a gas over water, there is water vapor mixed in with the gas. </li></ul><ul><li>To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure. </li></ul>
  48. 56. Mole Fraction For gas mixtures it is useful to define the mole fraction (X) X 1 = n 1 /n total The mole fraction or mole percent can be used to determine the partial pressures of gases in a mix IF the total pressure is known.
  49. 57. A container holds a mixture of oxygen, neon, and helium gases whose partial pressures are 150 torr, 300 torr, and 450 torr, respectively. The mole fraction of neon is <ul><li>0.17 </li></ul><ul><li>0.33 </li></ul><ul><li>0.50 </li></ul><ul><li>0.67 </li></ul>total i P P i  
  50. 58. Correct Answer: X i = P i / P total X i = (300 torr)/(150 + 300 + 450) torr X i = 300 torr/900 torr = 0.33 total i P P i   <ul><li>0.17 </li></ul><ul><li>0.33 </li></ul><ul><li>0.50 </li></ul><ul><li>0.67 </li></ul>
  51. 59. Section10.7 - 10.9 Homework due Thursday Gas Stoich Lab on Thursday Quiz 10 Friday Gas Stoich Lab due Monday
  52. 60. Ideal Gas LAW (PV=nRT) - A short statement that describes what happens (how gases behave) Kinetic Molecular THEORY - Explains why the gas laws are observed.
  53. 61. Kinetic-Molecular Theory <ul><li>This is a model that aids in our understanding of what happens to gas particles as environmental conditions change. </li></ul>
  54. 62. Main Tenets of Kinetic-Molecular Theory <ul><li>Gases consist of large numbers of molecules that are in continuous, random motion. </li></ul>
  55. 63. Main Tenets of Kinetic-Molecular Theory <ul><li>The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained. </li></ul><ul><li>Attractive and repulsive forces between gas molecules are negligible. </li></ul>
  56. 64. Main Tenets of Kinetic-Molecular Theory <ul><li>Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant. </li></ul>
  57. 65. Main Tenets of Kinetic-Molecular Theory <ul><li>The average kinetic energy of the molecules is proportional to the absolute temperature. </li></ul><ul><li>At any given temperature the molecules of all gases have the same average kinetic energy. </li></ul>
  58. 67. <ul><li>HCl (298 K) and H 2 (298 K) have the same average kinetic energy while O 2 (350 K) has a higher average kinetic energy. </li></ul>
  59. 68. Gas pressure is caused by gas molecules colliding with a surface.
  60. 69. Gas pressure is caused by gas molecules colliding with a surface. Gas Pressure increases when the frequency of collisions or the force of collisions increases.
  61. 70. Gas pressure is caused by gas molecules colliding with a surface. Gas Pressure increases when the frequency of collisions or the force of collisions increases. Collision frequency increases when temperature increases because molecules move faster.
  62. 71. Gas pressure is caused by gas molecules colliding with a surface. Gas Pressure increases when the frequency of collisions or the force of collisions increases. Collision frequency increases when temperature increases because molecules move faster. Collision frequency also increases when the amount of gas is increased or when the volume of a gas is decreased.
  63. 72. Gas pressure is caused by gas molecules colliding with a surface. Gas Pressure increases when the frequency of collisions or the force of collisions increases. Collision frequency increases when temperature increases because molecules move faster. Collision frequency also increases when the amount of gas is increased or when the volume of a gas is decreased. The force of collisions is increased with rising temperature.
  64. 73. Using KMT to explain gas laws Avogadro’s- More particles make more collisions, increasing pressure.
  65. 74. Using KMT to explain gas laws Avogadro’s- More particles make more collisions, increasing pressure. Boyle’s- More volume means less collisions because molecules must travel farther. Less collisions create less pressure.
  66. 75. Using KMT to explain gas laws Avogadro’s- More particles make more collisions, increasing pressure. Boyle’s- More volume means less collisions because molecules must travel farther. Less collisions create less pressure. Gay-Lussac’s- Higher temp means faster moving particles. Particles collide more often and with more force. More collisions create more pressure.
  67. 76. Using KMT to explain gas laws Avogadro’s- More particles make more collisions, increasing pressure. Boyle’s- More volume means less collisions because molecules must travel farther. Less collisions create less pressure. Gay-Lussac’s- Higher temp means faster moving particles. Particles collide more often and with more force. More collisions create more pressure. Charles’- Higher temp means faster moving particles. Particles collide more often and with more force. To maintain constant pressure, volume must increase.
  68. 77. A sample of He gas initially at STP is compressed to a smaller volume at constant temperature. What effect does this have on the rms speed of the atoms? <ul><li>Increases </li></ul><ul><li>Decreases </li></ul><ul><li>No effect </li></ul>
  69. 78. Correct Answer: The rms speed is directly proportional to the square root of the temperature, which does not change in this example. <ul><li>Increases </li></ul><ul><li>Decreases </li></ul><ul><li>No effect </li></ul>
  70. 79. KE = ½ m v 2 At the same temperature two gases have equal KE.
  71. 80. KE = ½ m v 2 At the same temperature two gases have equal KE. Heavier gases must have slower velocity to keep KE constant at a given temperature. Heavier gases move slower
  72. 81. Arrange the gases according to increasing molecular speed. <ul><li>He (25) < He (100) < Ne (25) < Ne (0) </li></ul><ul><li>He (25) < He (100) < Ne (0) < Ne (25) </li></ul><ul><li>Ne (0) < Ne (25) < He (25) < He (100) </li></ul><ul><li>Ne (25) < Ne (0) < He (100) < He (25) </li></ul><ul><li>Ne (0) < He (25) < Ne (25) < He (100) </li></ul>He (25°C) He (100°C) Ne (25°C) Ne (0°C)
  73. 82. Arrange the gases according to increasing molecular speed. <ul><li>He (25) < He (100) < Ne (25) < Ne (0) </li></ul><ul><li>He (25) < He (100) < Ne (0) < Ne (25) </li></ul><ul><li>Ne (0) < Ne (25) < He (25) < He (100) </li></ul><ul><li>Ne (25) < Ne (0) < He (100) < He (25) </li></ul><ul><li>Ne (0) < He (25) < Ne (25) < He (100) </li></ul>He (25°C) He (100°C) Ne (25°C) Ne (0°C)
  74. 83. Effusion <ul><li>The escape of gas molecules through a tiny hole into an evacuated space. </li></ul>
  75. 84. Effusion <ul><li>The escape of gas molecules through a tiny hole into an evacuated space. </li></ul>Compare effusion rates of two gases r 1 /r 2 =  M 2 /M 1 r = rate of effusion M = molar mass
  76. 85. An unknown gas effuses at half the rate of helium. This gas is likely to be which of the following? <ul><li>H 2 </li></ul><ul><li>CH 4 </li></ul><ul><li>Ne </li></ul><ul><li>O 2 </li></ul><ul><li>Ar </li></ul>
  77. 86. Correct Answer: ( r 1 /r 2 ) 2 = M 2 /M 1 M 2 = ( r 1 / r 2 ) 2 M 1 M 2 = (2/1) 2 (4.0 g/mol) = 16.0 g/mol Therefore it could be CH 4 <ul><li>H 2 </li></ul><ul><li>CH 4 </li></ul><ul><li>Ne </li></ul><ul><li>O 2 </li></ul><ul><li>Ar </li></ul>
  78. 87. Diffusion <ul><li>The spread of one substance throughout a second substance. </li></ul><ul><li>Heavier gases diffuse slower than gases of lower molar mass. </li></ul>
  79. 88. Diffusion occurs slower than effusion because the mean free path of molecules is reduced. Diffusion
  80. 90. <ul><li>Increasing pressure decreases mean free path. </li></ul>
  81. 91. <ul><li>Increasing pressure decreases mean free path. </li></ul>Increasing temperature has no effect on mean free path.
  82. 92. SAMPLE EXERCISE 10.15 Applying Graham’s Law An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is only 0.355 times that of O 2 at the same temperature. Calculate the molar mass of the unknown, and identify it. We can use Graham’s law of effusion to determine the molar mass of the unknown gas. If we let r x and represent the rate of effusion and molar mass of the unknown gas, Graham’s Law can be written as follows: From the information given, Thus,
  83. 93. Because we are told that the unknown gas is composed of homonuclear diatomic molecules, it must be an element. The molar mass must represent twice the atomic weight of the atoms in the unknown gas. We conclude that the unknown gas is I 2 . SAMPLE EXERCISE 10.15 continued We now solve for the unknown molar mass,
  84. 94. Real Gases <ul><li>In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure (<10atm). </li></ul>
  85. 95. Real Gases <ul><li>In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure (<10atm). </li></ul>
  86. 96. Deviations from Ideal Behavior <ul><li>The assumptions made in the kinetic-molecular model break down at high pressure and/or low temperature. </li></ul>
  87. 97. Deviations from Ideal Behavior <ul><li>Gas particles do have volume and this becomes significant at high pressures. The IGL underestimates the volume of gases. </li></ul>
  88. 98. Deviations from Ideal Behavior <ul><li>Gas particles are attracted to each other which causes the actual pressure to be less than what the IGL predicts. </li></ul>
  89. 99. Deviations from Ideal Behavior <ul><li>At high temperatures these effects are diminished as fast moving particles are less effected by intermolecular forces. </li></ul>
  90. 100. Deviations from Ideal Behavior <ul><li>The Ideal Gas Law accurately describes gases at low pressure and high temperatures. </li></ul>
  91. 102. <ul><li>100 K and 5 atm </li></ul><ul><li>Gases deviate the most from ideal behavior at low temp & high pressure. </li></ul>
  92. 103. Ideal Gases <ul><li>Gases with small size and low intermolecular forces behave most ideally. </li></ul>
  93. 104. Ideal Gases <ul><li>Gases with small size and low intermolecular forces behave most ideally. </li></ul><ul><li>The larger the gas molecule, and the stronger the IMF, the more deviation from ideal behavior. </li></ul>
  94. 106. <ul><li>Gases deviate from ideal behavior because the molecules have finite sizes and there are some attractions between the molecules. </li></ul>
  95. 107. Corrections for Nonideal Behavior <ul><li>The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account. </li></ul><ul><li>The corrected ideal-gas equation is known as the van der Waals equation . </li></ul>
  96. 108. The van der Waals Equation ) ( V − nb ) = nRT n 2 a V 2 ( P +

×